Physics-Interference

NEET Physics Interference MCQ Question

Type: MCQ-numerical-Easy-Class 12

In Young's Double Slit Experiment, if the slit separation dd is 0.025 mm, the distance between the slits and the screen DD is 5 cm, and the wavelength of light λ\lambda is 5 \times 10^{-5} cm, what is the distance xx from the central maximum to the first bright fringe?

A

0.1 mm

B

0.5 mm

C

0.025 mm

D

1 mm

Correct Answer

Option A

Detailed Explanation

Using the formula x=nDλdx = n \frac{D\lambda}{d} for constructive interference with n=1n = 1, the distance xx is calculated as 5×105×50.025=0.1\frac{5 \times 10^{-5} \times 5}{0.025} = 0.1 mm.

Found an issue with this question?